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286. Walls and Gates πŸ”’

Description

You are given an m x n grid rooms initialized with these three possible values.

  • -1 A wall or an obstacle.
  • 0 A gate.
  • INF Infinity means an empty room. We use the value 231 - 1 = 2147483647 to represent INF as you may assume that the distance to a gate is less than 2147483647.

Fill each empty room with the distance to its nearest gate. If it is impossible to reach a gate, it should be filled with INF.

 

Example 1:

Input: rooms = [[2147483647,-1,0,2147483647],[2147483647,2147483647,2147483647,-1],[2147483647,-1,2147483647,-1],[0,-1,2147483647,2147483647]]
Output: [[3,-1,0,1],[2,2,1,-1],[1,-1,2,-1],[0,-1,3,4]]

Example 2:

Input: rooms = [[-1]]
Output: [[-1]]

 

Constraints:

  • m == rooms.length
  • n == rooms[i].length
  • 1 <= m, n <= 250
  • rooms[i][j] is -1, 0, or 231 - 1.

Solutions

Solution 1

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class Solution:
    def wallsAndGates(self, rooms: List[List[int]]) -> None:
        """
        Do not return anything, modify rooms in-place instead.
        """
        m, n = len(rooms), len(rooms[0])
        inf = 2**31 - 1
        q = deque([(i, j) for i in range(m) for j in range(n) if rooms[i][j] == 0])
        d = 0
        while q:
            d += 1
            for _ in range(len(q)):
                i, j = q.popleft()
                for a, b in [[0, 1], [0, -1], [1, 0], [-1, 0]]:
                    x, y = i + a, j + b
                    if 0 <= x < m and 0 <= y < n and rooms[x][y] == inf:
                        rooms[x][y] = d
                        q.append((x, y))

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class Solution {
    public void wallsAndGates(int[][] rooms) {
        int m = rooms.length;
        int n = rooms[0].length;
        Deque<int[]> q = new LinkedList<>();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (rooms[i][j] == 0) {
                    q.offer(new int[] {i, j});
                }
            }
        }
        int d = 0;
        int[] dirs = {-1, 0, 1, 0, -1};
        while (!q.isEmpty()) {
            ++d;
            for (int i = q.size(); i > 0; --i) {
                int[] p = q.poll();
                for (int j = 0; j < 4; ++j) {
                    int x = p[0] + dirs[j];
                    int y = p[1] + dirs[j + 1];
                    if (x >= 0 && x < m && y >= 0 && y < n && rooms[x][y] == Integer.MAX_VALUE) {
                        rooms[x][y] = d;
                        q.offer(new int[] {x, y});
                    }
                }
            }
        }
    }
}

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class Solution {
public:
    void wallsAndGates(vector<vector<int>>& rooms) {
        int m = rooms.size();
        int n = rooms[0].size();
        queue<pair<int, int>> q;
        for (int i = 0; i < m; ++i)
            for (int j = 0; j < n; ++j)
                if (rooms[i][j] == 0